A substance has a half-life of 72 hours. In how many days will the substance have about 10%?

Question

A substance has a half-life of 72 hours (3 days).
In about how many days will the substance have about 10% of its original amount remaining?
Please explain to me how to do this problem.

Answer 1

1/2 the amount = (entire amount) x (growth factor)^72

take log of both sides:
log (.5) = (1) x 72 (log G)
-.3010 = 72 log(G)
-.00418 = log (G)
G=.9904 Thats the growth factor.

NOW: 10%= 0.10 so:

0.10 = (1)(.9904)^t
log .10 = t x (log .9904)
log .10/ (log .9904) = t
t = 238.7 hours
t= 9.945 days

Answer 2

In about 9-10 days.

Answer 3

Around 9-10 days.

Answer 4

ok, purely set up your equation like this. A = Ao*(a million/2)^(T) the place: A = volume of stuff last Ao = unique volume of stuff previously decaying began T = form of 0.5-lives that have exceeded (3-day sessions) #a million whilst 10% of the unique volume is last A = 0.a million*Ao. So: 0.a million*Ao = Ao*(a million/2)^(T) 0.a million = (a million/2)^T #2 whilst 70% of the unique volume is last A = 0.7*Ao. So: 0.7*Ao = Ao*(a million/2)^(T) 0.7 = (a million/2)^(T) For the two issues, purely take the log base (a million/2) of the two area. This makes the (a million/2)^(T) on the right area replace into T (think of related to the definition of a logarithm a at the same time as and you will comprehend why). Then use the replace of base formulation to discover the logs with base (a million/2) utilizing your calculator on the left area. eg: ln(.a million)/ln(a million/2) Now you have solved for T (form of 0.5-lives). to discover the form of days, purely undergo in concepts that there are 3 days consistent with 0.5 life.

Answer 5

If you are given the half-life of a problem you can find the decay constant.

Let A be the amount of substance at time t=0, then after a half life time had elapsed you would have 1/2 A left at time t=half_life

(1/2)A = A*e^(-k*half_life)

1/2 = e^(-k*half_life), take the natural log of each side
-[ln(1/2)]/(half_life) = k

In general k = 0.6931/(half_life) then 0.6931/3days = 0.2310/day

ln( 0.10) = (-0.2310*t), solve to t, t is 9.96 days round to 10 days.

In general the formula for the function of amout present at time t is

A(t) = A(t=o)*e(-k*t)